Chapter Solutions Radiation Sources � Problem 1.1 Radiation Energy Spectra: Line vs Continuous Line (or discrete energy): a, c, d, e, f, and i Continuous energy: b, g, and h � Problem 1.2 Conversion electron energies compared Since the electrons in outer shells are bound less tightly than those in closer shells, conversion electrons from outer shells will have greater emerging energies Thus, the M shell electron will emerge with greater energy than a K or L shell electron � Problem 1.3 Nuclear decay and predicted energies We write the conservation of energy and momentum equations and solve them for the energy of the alpha particle Momentum is given the symbol "p", and energy is "E" For the subscripts, "al" stands for alpha, while "b" denotes the daughter nucleus pal pal � pb � mal � Eal pb 2 mb � Eb Eal � Eb � Q and Q � 5.5 MeV Solving our system of equations for Eal, Eb , pal, pb , we get the solutions shown below Note that we have two possible sets of solutions (this does not effect the final result) Eb � 5.5 � mal Eal � mal � mb 3.31662 pal � � mal mb 5.5 mal mal � mb 3.31662 pb � � mal � mb mal mb mal � mb We are interested in finding the energy of the alpha particle in this problem, and since we know the mass of the alpha particle and the daughter nucleus, the result is easily found By substituting our known values of mal � and mb � 206 into our derived Ealequation we get: Eal � 5.395 MeV Note : We can obtain solutions for all the variables by substituting mb � 206 and mal � into the derived equations above : Eal � 5.395 MeV Eb � 0.105 MeV pal � � 6.570 amu � MeV pb � � 6.570 amu � MeV � Problem 1.4 Calculation of Wavelength from Energy Since an x-ray must essentially be created by the de-excitation of a single electron, the maximum energy of an x-ray emitted in a tube operating at a potential of 195 kV must be 195 keV Therefore, we can use the equation E=h�, which is also E=hc/Λ, or Λ=hc/E Plugging in our maximum energy value into this equation gives the minimum x-ray wavelength Λ� h�c E where we substitute h � 6.626 � 10�34 J � s, c � 299 792 458 m � s and E � 195 keV Chapter Solutions Λ � 1.01869 J–m � 0.0636 Angstroms KeV � Problem 1.5 235 UFission Energy Release Using the reaction 235 U � 117 Sn � 118 Sn, and mass values, we calculate the mass defect of: M �235 U� � �M �117 Sn� � M �118 Sn�� � �M and an expected energy release of �Mc2 Q � �235.0439 � �116.9029 � 117.9016�� AMU � 931.5 MeV � 223 MeV AMU This is one of the most exothermic reactions available to us This is one reason why, of course, nuclear power from uranium fission is so attractive � Problem 1.6 Specific Activity of Tritium Here, we use the text equation Specific Activity = (ln(2)*Av)/ �T1�2 *M), where Av is Avogadro's number, T1�2 is the half-life of the isotope, and M is the molecular weight of the sample Specific Activity � ln�2� � Avogadro ' s Constant T1�2 M We substitute T1�2 � 12.26 years and M= Specific Activity � grams mole to get the specific activity in disintegrations/(gram–year) 1.13492 � 1022 gram –year The same result expressed in terms of kCi/g is shown below Specific Activity � 9.73 kCi gram � Problem 1.7 Accelerated particle energy The energy of a particle with charge q falling through a potential �V is q�V Since �V= MV is our maximum potential difference, the maximum energy of an alpha particle here is q*(3 MV), where q is the charge of the alpha particle (+2) The maximum alpha particle energy expressed in MeV is thus: Energy � Mega Volts � Electron Charges � MeV Chapter Solutions � Problem 1.8 Photofission of deuterium The reaction of interest is D� 0 Γ � 10 n � 1 D�Γ� n� 1 p + Q (-2.226 MeV) p+ Q (-2.226 MeV) Thus, the Γ must bring an energy of at least 2.226 MeV in order for this endothermic reaction to proceed Interestingly, the opposite reaction will be exothermic, and one can expect to find 2.226 MeV gamma rays in the environment from stray neutrons being absorbed by hydrogen nuclei � Problem 1.9 Neutron energy from D-T reaction by 150 keV deuterons We write down the conservation of energy and momentum equations, and solve them for the desired energies by eliminating the momenta In this solution, "a" represents the alpha particle, "n" represents the neutron, and "d" represents the deuteron (and, as before, "p" represents momentum, "E" represents energy, and "Q" represents the Q-value of the reaction) pa pa � pn � pd ma pn � Ea mn � En pd 2 md � Ed Ea � En � Ed � Q Next we want to solve the above equations for the unknown energies by eliminating the momenta (Note : Using computer software such as Mathematica is helpful for painlessly solving these equations) We evaluate the solution by plugging in the values for particle masses (we use approximate values of "ma ," "mn ,"and "md " in AMU, which is okay because we are interested in obtaining an energy value at the end) We define all energies in units of MeV, namely the Q-value, and the given energy of the deuteron (both energy values are in MeV) So we substitute ma = 4, mn = 1, md = 2, Q = 17.6, Ed = 0.15 into our momenta independent equations This yields two possible sets of solutions for the energies (in MeV) One corresponds to the neutron moving in the forward direction, which is of interest En � 13.340 MeV En � 14.988 MeV Ea � 4.410 MeV Ea � 2.762 MeV Next we solve for the momenta by eliminating the energies When we substitute ma = 4, mn = 1, md = 2, Q = 17.6, Ed = 0.15 into these equations we get the following results pn � pd � pd � 352 pa � 10 pd � 2 pd � 352 We know the initial momentum of the deuteron, however, since we know its energy We can further evaluate our solutions for pn and pa by substituting: pd � � � 0.15 The particle momenta ( in units of pn � �5.165 pn � 5.475 amu� MeV ) for each set of solutions is thus: pa � 5.940 pa � �4.700 The largest neutron momentum occurs in the forward (+) direction, so the highest neutron energy of 14.98 MeV corresponds to this direction Chapter Solutions Radiation Interaction Problems � Problem 2.1 Stopping time in silicon and hydrogen Here, we apply Equation 2.3 from the text 1.2 range Tstop � mass energy 107 Now we evaluate our equation for an alpha particle stopped in silicon We obtained the value for "range" from Figure 2.8 (converting from mass thickness to distance in meters by dividing by the density of Si � 2330 mg� cm3 ) The value for "mass" is approximated as AMU for the alpha particle, and the value for "energy" is MeV We substitute �6 range � 22 x 10 , mass � and energy � into Equation 2.3 to get the approximate alpha stopping time (in seconds) in silicon Tstop � 2.361 � 10�12 seconds Now we the same for the same alpha particle stopped in hydrogen gas Again, we obtain the value for "range" (in meters) from Figure 2.8 in the same manner as before (density of H � 08988 mg� cm3 ), and, of course, the values for "mass" and "energy" are the same as before (nothing about the alpha particle has changed) We substitute range = 0.1, mass = and energy =5 into Equation 2.3 to get the approximate alpha stopping time (in seconds) in hydrogen gas Tstop � 1.073 � 10�8 seconds The results from this problem tell us that the stopping times for alphas range from about picoseconds in solids to nanoseconds in a gas � Problem 2.2 Partial energy lost in silicon for MeV protons �Clever technique: A MeV proton has a range of 210 microns in silicon according to Figure 2-7 So, after passing through 100 microns, the proton has enough energy left to go another 110 microns It takes about 3.1 MeV, according to the same figure, to go this 110 microns, so this must be the remaining energy Thus the proton must have lost 1.9 MeV in the first 100 microns � Problem 2.3 Energy loss of MeV alpha in microns Au From Figure 2.10, we find that �1 dE � Ρ dx 380 MeV�cm2 g Therefore, dE � dx 380 MeV�cm2 * g Ρ (ignoring the negative sign will not affect the result of this problem) Energy loss � We substitute dE �dx � (in non-SI units) Ρ �dE � dx� �x Ρ 380 MeV cm2 Ρ gram , Ρ� 19.32 grams cm3 and �x � microns to get the energy loss of the MeV Α-particle in Μm Au Chapter Solutions 36 708 MeV microns Energy loss � cm The result in SI units is thus: Energy loss � 3.671 � 106 eV Since this energy loss is greater than the initial energy of the particle, all of the Α-particle energy is lost before Μm Note the small range of the Α , i.e ~ Μm per MeV � Problem 2.4 Range of MeV electrons in Al Scaling Law The Bragg-Kleeman rule, or scaling law, allows us to relate the known range in one material to the range in another material The semi-empirical rule we use is: R1 R0 � Ρ0 A1 Ρ1 A0 (Equation 2.7) Here, we use Figure 2.14 to approximate the MeV electron range in silicon �R0 ), and since we know every other quantity in Equation 2.7, we can approximate the range of the MeV electron in aluminum �R1 ) Solving for R1 we can find the estimated range of the electron in Aluminum AWAl AWSi rSi ΡSi rAl � We substitute rSi � ΡAl 0.5 g cm2 ΡSi , AWAl � 26.9815 amu, AWSi � 28.0855 amu, ΡAl � 2.698 g cm3 and ΡSi � 2.329 g cm3 to get the approximate range of MeV electrons in aluminum (in cm) rAl � 0.1816 cm � Problem 2.5 Compton scattering This problem asks for the energy of the scattered photon from a MeV photon that scattered through 90 degrees We use the Compton scattering formula (Equation 2.17) We write the Compton scattering formula, defining the scattering angle ("Θ") as 90 degrees and the photon energy ("E0 ") as MeV Energy � E0 �1�Cos�Θ�� E0 me � c2 �1 We substitute Θ = 90° and E0 � MeV to get the energy of the scattered photon in MeV Energy � 0.338 MeV � Problem 2.6 Prob of photoelectric in Si versus Ge Chapter Solutions Problem 2.6 Prob of photoelectric in Si versus Ge For a rough estimate, we can note that photoelectric probabilities vary as � Z 4.5 so we would expect that ΤSi � ΤGe � � 14 � 32� 4.5 � 0.0242 � Problem 2.7 The dominant gamma ray interaction mechanism See Figure 2-20 and read off the answers (using the given gamma-ray energies and the Z-number for the given absorber in each part) Compton scattering: a, b, and d Photoelectric absorption: c Pair production: e � Problem 2.8 Mean free path in NaI of MeV gamma-rays (a) The gamma-ray mean free path (Λ) in NaI is � Μ ( where Μ is the total linear attenuation coefficient in NaI) The mass attenuation coefficient ( ) is 0.06 cm2 � gm at MeV according to Figure 2.18, and the density of NaI relative to water (Ρ) is Μ Ρ 3.67 gm� cm3 (by the definition of specific gravity) Therefore, we have Λ = � Μ = Μ Ρ �Ρ Here, we will denote the mass Μ attenuation coefficient ( ) by Μ Ρ , so we have Ρ Λ� �Μ Ρ Ρ� We substitute Μ p � 0.06 cm2 g and Ρ � 3.67 g cm3 to get the mean free path of MeV gamma-rays in NaI (in cm) Λ � 4.54 cm (b) Any photon which emerges from cm cannot have undergone a photoelectric absorption Neglecting buildup factors, the probability that a photon emerges from the slab without having an interaction is e�ΜT x , where ΜT is the total attenuation coeffiinteraction (1-e�ΜT x ) The probability that the interaction is a photoelectric interaction is Τ � ΜT (this is not the probability per unit cient The complement of this is the probability that a photon doesn't emerge from the slab without having had at least one photon undergoes photoelectric absorption in the slab is (Τ � ΜT )*(1-e�ΜT x ) This equation is expressed below, along with the path length, but the total probability that any given interaction is a photoelectric interaction) Therefore, the probability that a values for ΜT (which is just divided by the previous result for Λ), the attenuation distance (denoted "x" and which is cm), and Τ, which is just the mass attenuation coefficient for photoelectric absorption (found on Figure 2.18 to be 0.01) multiplied by the density of NaI (3.67 g/cm3) Probability of photoelectric absorption � We substitute ΜΤ � � cm NaI , 4.54 cm x � cm and Τ � 0.01�3.67 cm Τ �1 � ��ΜΤ x � Μ to get the probability of photoelectric absorption for 600 keV gamma-rays in Chapter Solutions We substitute ΜΤ � , 4.54 cm x � cm and Τ � 0.01�3.67 cm to get the probability of photoelectric absorption for 600 keV gamma-rays in cm NaI Probability of photoelectric absorption � 0.0329 What is interesting is that a different result is obtained using a different, although seemingly equally valid approach We can note that the probability per unit path length of a photoelectric interaction is Τ, so � e�Τx is the probability of a photoelectric interaction in traveling a distance x Probability of a photoelectric interaction � � ��Τ x We substitute x=1cm and Τ= 0.010�3.67 to cm get the probability of a photoelectric interaction in traveling a distance x Probability of a photoelectric interaction � 0.0360 This result is slightly (10%) larger from the previous answer because this approach does not account for the attenuation of photons through the material by other means � Problem 2.9 Definitions See text � Problem 2.10 Mass attenuation coefficient for compounds The linear attenuation coefficient is the probability per path length of an interaction In a compound, the total linear attenuation coefficient would be given by the sum over the ith constituents multiplied by the density of the compound: Μc = Ρc �wi � Ρ � Μ (Equation 2.23) i where wi is the weight fraction of the ith constituent in the compound (represented by "wH " and "wO ," respectively, in this problem), �Μ � Ρ�i is the mass attenuation coefficient of the ith constituent in the compound (represented by "ΜH " and "ΜO ," respec- tively), and Ρc is the density of the compound (represented by "ΡW " for the density of water) The expression below shows this sum for water attenuating 140 keV gamma rays: Μc � ΡW �ΜH wH � ΜO wO � We substitute ΜH � 26 cm2 � g, ΜO � 14 cm2 � g, ΡW � g � cm3 , wH � � 18 and wO � 16 �18 to get the linear attenuation coefficient for water attenuating 140 keV gamma rays Μc � 0.153333 cm The mean free path, Λ, is just the inverse of this last calculated value, or The mean free path of 140 keV gamma rays in ΜC water is thus: ΛH2 O � 6.52 cm This turns out to be an important result because Tc � 99 m is a radioisotope routinely used in medical diagnostics and it emits 140 keV gamma rays Since most of the human body is made of water, this gives us an idea of how far these gamma rays can travel without an interaction through the human body and into our detectors Chapter Solutions This turns out to be an important result because Tc � 99 m is a radioisotope routinely used in medical diagnostics and it emits 140 keV gamma rays Since most of the human body is made of water, this gives us an idea of how far these gamma rays can travel without an interaction through the human body and into our detectors � Problem 2.11 J of energy from MeV depositions We are looking for the number of MeV alpha particles that would be required to deposit J of energy, which is the same as looking for how many MeV energy depositions equal J of energy We expect the number to be large since J is a macroscopic unit of energy To find this number, we simply take the ratio between J and MeV, noting that MeV = 1.6 x 10�13 J The number of MeV alpha particles required to deposit J of energy is thus: n� Joule � 1.25 � 1012 alpha particles MeV � Problem 2.12 Beam energy deposition 100 ΜA tells us the current, i.e., the number of Coulombs passing in unit time If we divide this by the amount of charge per particle, it gives us the number of particles passing through the area in unit time If we take this value, and multiply it by the energy per particle, we get the energy/time, or power The power dissipated in the target by a beam of MeV electrons with a current of 100 ΜA is thus: Power � �100 ΜA� �1 MeV� �Electron Charge� � 100 Watts This is a surprisingly low amount of power, about the same as from a light bulb, and represents the output from a small accelerating device � Problem 2.13 Exposure rate m from Ci of 60 Co We use the following equation for exposure rate: Exposure Rate = �s Α d2 (Equation 2.31) where Α is the source activity, d is the distance away from the source, and �s is the exposure rate constant The exposure rate constant for Co-60 is 13.2 R � cm2 � hr � mCi (from Table 2.1) The exposure rate m from a Ci Co-60 source is thus: Exposure Rate � �s Exposure Rate � Α d2 where we substitute Α � Ci, d � m, �s � 13.2 0.528 cm2 R hr mm2 � 52.8 mR The result in SI units : Exposure Rate � 3.78 � 10�9 C kg � s � Problem 2.14 �T/�t from 10 mrad/hr hr R � cm2 hr � mCi Chapter Solutions Problem 2.14 �T/�t from 10 mrad/hr The dose of rad corresponds to an energy deposition of 100 ergs/gram So 10 mrad/hr corresponds to erg/(gram-hr) Using a specific heat of water as calorie � �gram �o C), we can use the equation �Q � mC p �T Our given ergs/(gram-hr) is �Q/(m*�t) If we divide our equation by mC p , we are left with �T/�t on the right side, which is the quantity of interest, i.e �T/�t = �Q��m�t� Cp �T �t � = ergs��gram�hr� calorie��gram �o C� Erg Gram Calorie Hour Gram Centigrade The result below is the rate of temperature rise in a sample of liquid water with an absorbed dose rate of 10 mrad/h Note that this is virtually impossible to measure because it is so small �T �t � 2.39 � 10�8 Centigrade Hour There are unusual radiation detectors which actually use the temperature rise in a detecting material to detect ionizing radiation For the curious reader, some research on bolometers in radiation detection, and also read about the superconducting radiation detectors under development � Problem 2.15 Fluence-dose calculations for fast neutron source The Cf source emits fast neutrons with the spectrum N(E) dE given in the text by Eqn 1.6 Each of those neutrons carries a dose h(E) that depends on its energy as shown in Fig 2.22(b) To get the total dose, we have to integrate N(E)h(E) over the energy range First, let's consider only the neutron dose h(E) In the MeV range, we need a linear fit to the log h-log E plot of Figure 2.22(b) by using two values read off the curve: Log10 �10�12 � � b � m Log10 �0.01� Log10 �10�10 � � b � m Log10 �1.0� Solving the system of equations above yields : m � and b � �10 This result gives us an approximate functional form for h(E) �in Sv � cm2 �= 10�10 E �MeV� Recall Sv = 100 Rem = 105 mrem In a moment, we'll integrate N(E)h(E) to get the total dose-area, but first check the normalization of N(E): norm � � � E � � E 1.3 �E � 1.31359 We'll need this normalization factor because we will want to use N(E)/norm as the probability that a source neutron has energy E Now, a source neutron � cm2 arriving at the person delivers a dose (in Sv) of: �0 E3�2 � � neutron � dose � 10 � E 1.3 �E � 1.95 � 10�10 Sv 10 norm �We now need to multiply this by the number of neutrons produced by micrograms of Cf-252 (2.3 x 106 n/sec-mg) at meters over hours: Chapter Solutions We now need to multiply this by the number of neutrons produced by micrograms of Cf-252 (2.3 x 106 n/sec-mg) at meters over hours: dose � equivalent � 2.3 � 106 �3 Μg� �8 � 3600 sec� Μ g sec� Π 5002 � � 63 254.5 neutrons � cm2 So the total dose equivalent is given by (using 105 mrem/Sv): dose equivalentneutrons � 1.23 mrem This is a small dose, comparable to natural background Aside: What about the dose from the gamma rays? They are high-energy gammas and the source emits 9.7 gammas per fission Using a value of hE ~ x 10�12 Sv � cm2 : dose equivalent Γ � �2.3 � 106 neutrons� �3 Μg� �8 � 3600 sec� �5 Sv cm2 � �100 Rad� �1000 mRad� �1 Μg sec� �1012 neutrons� �4 Π 500 cm2 � �Sv Rad� � 0.0316 mRad So the gamma dose is even smaller than that from the fast neutrons, as expected Fast neutrons have a high quality factor, i.e., they produce a heavy charged particle when they interact, and therefore a lot more biological damage than the light electrons produced when gamma rays interact in materials 10 Chapter 17 Solutions � Problem 17.19 Improving true to chance coincidence ratio In answering these questions, recall that the chance coincidence rate is rtrue = S Ε1 Ε2 and rchance =( r1 r2 Τ ) = (S Ε1 S Ε2 Τ) so rtr rch � SΤ a Changing the solid angle does not change S or Τ so there is no effect b Increasing S reduces the true to chance coincidence ratio as shown in the equation above c Increasing Τ reduces the ratio above d Increasing the energy window slightly only changes the efficiency by which the pulses are selected If one opens this window too large so that pulses other than the desired energy are included, then additional pulses that are not coincidence pulses are allowed, thereby effecively increasing the number of non-true coincidence pulses and thereby decreasing the ratio � Problem 17.20 Interpretting a Coincidence Delay curve experiment (a) The width of the step is 2Τ, so Τ � 125�25 � 50 ns (b) The full width at the base of the prompt coincidence peak in the time spectrum is what gives rise to the sloped edges on the coincidence delay curve, and is 140-110=40-10= 30 ns (c) Assuming the singles rates are the same, i.e., r1 � r2 � rs , then using rchance � rs Τ, and noting from the curve that the background is the chance rate rch � cps, we solve for rs : rs � rchance 2Τ � 7071 � s � Problem 17.21 Subranging ADC A 12 bit-subranging ADC can consist of stages of bits or stages of bits Which has the smallest (a) number of comparators, and (b) latency (assuming the same clock frequency)? The number of comparitors needed is related to the number of bits of information N according to 2N So the 3-stages of four bit ADCs use � 24 = 48 comparators, whereas stages of three bit ADCs use � 23 = 32 comparators While the stage ADC needs more comparators, the latency depends on the number of stages, however, so the 4-stage ADC can be expected to have a longer latency However, since the architecture involves serial*pipelining, the throughput should be proportional to the clock speed which we assumed was constant and should be the same for both cases Looking at the two extremes of the subranging ADC, an N-bit subranging ADC which has: (a) stage is known as a Flash ADC, and (b) N stages is a successive approximation ADC � Problem 17.22 Digital Filtering For a step function with Vi =1 for i � 0, and elsewhere, apply four different digital filters We will use Matematica for this problem We define our input function: : 127 Chapter 17 Solutions We define our input function: : V�i_� :� for i � V�i_� :� for i � Our first filter is: h � �0.25, 0.25, 0.25, 0.25� The filter output is just the dot product of the filter and the input vectors over the length of the filter The output looks like : �0., 0., 0.25, 0.5, 0.75, 1., 1., 1., 1., 1., 1.� which, when plotted, looks like: 0.8 0.6 0.4 0.2 10 Changing the filter code to that of a "derivative" and reversing the code order: h � �1, � 1, 0, 0� yields a filtered output of: 0.8 0.6 0.4 0.2 (c) Here is a "differentiator" filter with memory: h � �0.5, 0.5, � 0.5, � 0.5� and this is the output of the filter applied to the input: 128 10 Chapter 17 Solutions 0.8 0.6 0.4 0.2 10 (d) Finally, here is the exponential integrating filter: h � �1, 1 , � � , �3�2 , �2 � which produces the outcome: 1.5 0.5 10 129 Chapter 18 Solutions Multichannel Pulse Analysis � Problem 18.1 Channels needed for fixed detector resolution For a fixed resolution, we want the peak FWHM to contain channels, so FWHM = channels = (Resolution)*(Peak centroid channel) The ADC must have at least the Peak centroid channel number of channels, so the ADC must have: ADC � � 1667 channels 003 � Problem 18.2 Channels versus gain We note that the channel a pulse amplitude is mapped to will be linearly related to the gain, where k is the proportionality constant: Channel = k (Pulse Amplitude) (Gain) + Zero Offset The separation between two centroids is thus given by: �Channel = k (� Pulse Amplitude) (Gain) If the detector pulse amplitudes are kept constant, but the gain is changed, we find that the separation is changed as: ∆��Channel� �Channel ∆�Gain� Gain = change in separation � �250 � 24 � �6 channels 1000 So the centroid channel separation decreases by channels � Problem 18.3 Wilkinson ADC Oscillator Frequency We have to be able to resolve the maximum amplitude into 2048 parts in 25 Μs Each click of the clock corresponds to one channel number The conversion time is given by tc � If the offset is zero, then: f � PH tc 130 � 2048 25 Μs � 81.9 MHz PH f where the Pulse Height PH refers to the pulse height channel number Chapter 18 Solutions � Problem 18.4 Successive Approx ADC steps for N channels Recall that each ADC comparison gives us bit of information Suppose our spectrum contains M channels That means we must have a measurement accuracy equal to 1/M Letting N be the number of ADC steps (comparisons), we must have 2N � M , so N= Log[M]/Log[2]: N� ln�M� � 12 ADC steps where we substituted M � 4096 ln�2� � Problem 18.5 Deadtime losses in Wilkinson ADC If the ADC uses a linear ramp, then the deadtime is proportional to the pulse-height and not a fixed value The expressions derived for the deadtime losses in Chapter assumed that the deadtime was a fixed value, regardless of the pulse height Note that other ADC designs, such as the Flash ADC or the Successive Approximation ADC have a fixed deadtime per conversion � Problem 18.6 Wilkinson ADC The ADC operates at 80 MHz and has a storage time of 2.5 Μs Since each clock cycle corresponds to one channel: (a) for channel 300, deadtime will be N/f + B dead time � 300 80 � 10 � 2.5 106 � 6.25 � 10�6 seconds (b) The fractional dead time for an event rate of r0 is given by mΤ where we assume a nonparalyzable model for m= n 1�nΤ Here we substitute n � fdead � nΤ nΤ� 5000 second and Τ � 2.5 Μs � 220 seconds 80�106 to get the fractional dead time as thus: � 0.0256 (c) Real time for a live time of 10 minutes in (b) above Here we substitute fdead � 0.025579 and live-time = 10 minutes to get the real time as thus: real time � live � time � fdead � 10.3 minutes � Problem 18.7 Periodic pulser dead time The key to this problem is to note that the pulser pulses are periodic Therefore, they will each be counted if the time between pulses is longer than the dead time per pulse �tdead � Noting the time between the pulses of frequency f is �t= 1/f, the count rate CR is less than f if �t � tdead or f < � tdead If the frequency is high enough, multiple pulses may be included in the dead time You can show that the count rate CR is given by: CR= f/(n+1) if f > �n � tdead � for n= {0,1,2, } A plot of CR versus f (in units of � tdead ) thus looks like a set of steps starting at height f and going down by half its height each step along the frequency axis � Problem 18.8 Multiparameter spectra 131 Chapter 18 Solutions Problem 18.8 Multiparameter spectra The number of memory channels required for two detector spectra will be D1 x D2 where the number of channels for either is given from Problem 18.1 as Di = (5 channels� � �Resolutioni � So the number of memory locations will be given by: locations � � R1 R2 We substitute R1 =.005 and R2 � 025 to get the number of memory locations locations � 200 000 memory locations � Problem 18.9 Channel counting statistics The average count in a channel should be zero, but this will fluctuate due to counting statistics Indeed, half of the channels will have negative values Since: N = (S+B)-(B) = B - B, then by the propagation of error: ΣN = Sqrt [2B] = ΣN � 600 � 24.5 counts So the average channel will have a value of 0±24.5 counts � Problem 18.10 Finding the Centroid Using the Gaussian function, let's look at the second derivative behavior Our function is: f �x� � y0 � � �x�x0 �2 Σ2 so the second derivative is given by (we substitute x0 � 0, Σ = and y0 � since these factors don't affect the results): �2 f �x� � x2 � � � x2 x2 � � � x2 � � � x2 �x2 � 1� We plot the function and its second derivative: f �x� 1.0 0.5 �3 �2 �2 f �x� �1 �0.5 � x2 �1.0 Note that the peak of the function is characterized by a large negative excursion in second derivative Hence, this is often used as the indicator of a peak in automated peak finding � 132 Chapter 18 Solutions Note that the peak of the function is characterized by a large negative excursion in second derivative Hence, this is often used as the indicator of a peak in automated peak finding � Problem 18.11* Peak Fitting of Data We are given an MCA data set consisting of a peak and constant background We first try to visually estimate the centroid, area, and FWHM The data in the list below corresponds to a channel number starting with channel 711, increasing by one, and ending with channel 728 Below we plot the data versus the channel number data � �238, 241, 219, 227, 242, 280, 409, 736, 1190, 1625, 1739, 1412, 901, 497, 308, 256, 219, 230� channel � �711, 712, 713, 714, 715, 716, 717, 718, 719, 720, 721, 722, 723, 724, 725, 726, 727, 728� 1500 1000 500 720 725 From the graph, we estimate a constant background of about 235 and FWHM of channels, with peak channel at 721 We estimate the area by summing the data values and then subtracting an assumed constant background over the 5th through 17th data points: area � �242 � 280 � 409 � 736 � 1190 � 1625 � 1739 � 1412 � 901 � 497 � 308 � 256 � 219� � �235 � 13� area � 6759 y�x�1� (b) Now, we try the linearized method described in the text We define ln[Q(x)] = ln[ y�x�1� ], a linear least squares fit of ln[Q(x)] versus x over channels, and plot this versus x (In this problem the channels we will use are channels 719 through 723) We first subtract off the background of 235 counts, and then proceed: data � raw counts � 235 � �3, 6, �16, �8, 7, 45, 174, 501, 955, 1390, 1504, 1177, 666, 262, 73, 21, �16, �5� We now plot x versus ln� data�i�1� � where i is the element number of the data list, starting with and ending at 13 data�i�1� 133 Chapter 18 Solutions 1.5 1.0 0.5 �0.5 �1.0 Now, let's a linear least squares fit over data points (9 � 13) We'll skip the weighting factor (data[[i-1]] data[[i+1]]/(data[[i-1]]+data[[i+1]])) here The (x, y) coordinates and the linear least squares fit line are shown below 1390 719 �log� 501 1504 720 �log� � 955 1390 721 log� 722 log� 723 log� � � y � 0.631443 x � 455.069 � 1177 752 333 1177 262 � We now find the centroid, FWHM and Σ x0 � �intercept �� slope Σ� �455.069 � 721 0.631443 � slope � 1.78 0.631443 Recognizing that the FWHM of a Gaussian is 2.35 time the standard deviation of the distribution: FWHM � 2.35 Σ � 2.35 � 4.18 channels 0.631443 Now let's get the area, assuming no uncertainties in the centroid and Σ: �13 i�9 data�i� area � 2Π Σ� �channel�i��x0 �2 Σ2 �13 i�9 data�i� �ln�data�i�� � 6788 This is quite close to our original estimate of the net area Extra: Note that we could also apply a nonlinear fit to the data using the three fitting parameters y0 , x0 , and Σ We find that a nonlinear fit for the data to the function y �x� � y0 � y0 � 1518, x0 � 721 and Σ � 1.78 and the net area is then: 134 � ��x�x0 �2 Σ2 � yields: Chapter 18 Solutions area � Σ y0 Π � 6774 The nonlinear fit is generally regarded as more accurate since it avoids the problem of a potential bias associated with biased estimators 135 Chapter 19 Solutions Miscellaneous Detectors � Problem 19.1 Cherenkov emission by electron (a) We use the energy threshold relationship for Cherenkov radiation: Energy � me c2 1� n2 � �1 where we substitute�n � 1.47 Energy � 0.186 MeV ing angle, with Ee � �2 EΓ � � �me � c2 � � EΓ � Solving this equation for EΓ and substituting Ee = 0.186173 MeV gives us: (b) For this amount of energy to be given to a Compton electron, the minimum gamma ray energy corresponds to a 180° scatter- EΓ � �0.144 MeV and EΓ � 0.330 MeV Clearly, the negative energy is not physical, so our solution is 330 keV � Problem 19.2 Fewer Cherenkov photons/MeV than in NaI Figure 19.2 shows about 600 photons/(2MeV e� ), or about 300 photons/MeV for Cherenkov emission in water In NaI, we expect about 38000 photons/MeV (see textbook) about 100 times larger light yield � Problem 19.3 Liquid Xe attenuation at 30 keV Figure 6.18 tells us that inches of gaseous Xe at atm will attenuate ~20% of 30 keV x-rays But the density of Xe gas at STP is 5.85 g/liter while the density of liquid Xe is 3.52 kg/liter Recall: I�t� � I0 = e��Μ�Ρ� �Ρt� � 80 � so the mass attenuation constant Μ/Ρ is ΜΡ � ΜΡ � Μ Ρ � �ln�.8� �Ρ t� and when we substitute Ρ � 5.85 gram � Liter and t � 5.08 cm 0.0075087 Liter gram–cm Using the previous expression, t = �1 � Ρ Μ Ρ � Log�I0 � I�t�� t� ln� � f �Ρ Μ Ρ � t � 262 Μm 136 where we substitute f � 0.5 and Ρ � 3.52 kg � Liter Chapter 19 Solutions � Problem 19.4 Intensifier screens for radiology X-rays have only a small probbilitiy of directly interacting in the radiographic film, so a high-Z scintillating material placed adjacent to the film to produce the light that exposes the emulsion In place of the film, one can also view the intensifier screen emissions by a CCD for real-time radiographic applications � Problem 19.5 Electron traps: Advantage or disadvantage? Radiation interactions in a crystaline material produce electrons and holes If the application requires an immediate readout of the interaction, then the electron traps are undesirable because the desired signal is held up by the electron being trapped at the impurity site For most scintillators used in spectroscopy, this is a disadvantage However, for dosimeters, where one wishes to integrate the total number of events over a period of time, then trapping the elctrons at an impurity site is an advantage The electrons are later released under thermal or optical stimulation, and the resulting emission measures the integrated number of events � Problem 19.6 Track detectors We are looking for a track etch detector that will be sensitive to fission products but not alpha particles Fission products are high Z (~100 amu) and high kinetic energy (~100 MeV) Alpha particles have A=4 and a typical kinetic energy of several MeV Looking at the referenced figure for He, any of the track materials whose horizontal lines are above that of Lexan will not produce a track density sufficient to register the alpha particle For the fission products, Cronar offers the highest sensitivty (lowest required ionization density to register a track) while still being insensitive to the alpha particles � Problem 19.7 Activation and counting (a) We need to find the induced activity for a 10 minute and 20 minute irradiation since the count rate is proportional to the induced activity Since the induced activity is given by A�t� � A� �1 � e�Λt �, the ratio we want is given by: A� � ��Λ t1 1�� �Λ t2 where we substitute Λ � ln�2� 2.3 , t1 � 10 and t2 � 20 �all time units are in minutes� A � 0.953 This says that a 10 minute irradiation gives 95% of the counts of a 20 minute irradiation (b) Suppose we counted forever (instead of just 10 minutes) What would the gain be in the number of net counts? Recall that Counts = k(1 - e�Λ tc ) and assume the irradiation and wait time remain constant So the factor increase in counts is given by: counts � 1 � ��Λ t where we substitute Λ � ln�2� and t � 10 2.3 counts � 1.05 Thus, only a 5% increase in counts would be obtained by increasing the counting time from 10 minutes to infinity This is because of the short 2.3 minute half life of this isotope it's mostly gone after 10 minutes 137 Chapter 19 Solutions � Problem 19.8 Role of Frisch Grid in HP Xe detectors A Frisch grid is employed to permit sensitivity to only one carrier type Since ions move so slowly in a gas and often don't make it to the cathode in a reasonable time, the Frisch grid permits the generation of the signal solely on the more mobile carrier since the detector signal is induced only when this fast carrier passes the grid and is collected on the anode Thus, the role of the other carrier plays no role in the signal formation and the output signal is proportional only to the number of fast carriers (electrons) collected 138 Chapter 20 Solutions Background and Detector Shielding � Problem 20.1 Background from 40 Kin NaI We note that the total number of NaI molecules in the detector will be ΡV*(Avogardro's Constant)/(Molecular Weight) where V= Πr2 Hfor a cylinder Natural potassium is 0.012% 40 K, which has a half life of 1.26 x 109 years We are allowed only cps in the detector volume So the maximum number of K atoms is Λ (0.012%)K = cps, or solving for K (where we are using the symbol K also for the number of potassium atoms): K� �1 � second� ln2 � 0.012 � � T1�2 where we substitute T1�2 � 1.26 � 109 years K � 4.78 � 1020 atoms But the total number of NaI molecules in the detector are: NaI � Ρ V �Avogadro ' s Constant� MolecularWeight where we substitute V � Π r2 H, MolecularWeight � 150 g mole , Ρ� 3.67 g cm3 ,r� 7.62 cm and H � 7.62 cm NaI � 5.12 � 1024 molecules So the atom ppm of K that is allowable is (noting atoms per NaI molecule): ppm � K � 106 � 46.7 atoms of K per million atoms NaI � Problem 20.2 14 C source When cosmic rays interact in the upper atmosphere, they produce a variety of elementary particles, including neutrons When one of these neutrons interacts with a nitrogen nucleus in the atmosphere, the (n,p) reaction 14 N �n, p� produces 14 C which then enters the environment and serves as chronometer When a living organism dies, it no longer ingests fresh carbon, and so the decreased relative abundance of the 14 C is a measure of the time since death � Problem 20.3 Cosmic ray pulse rejection Cosmic rays tend to be high energy charged particles that have a low probability of interacting in the relatively low density gas in the proportional tube (i.e., low dE ) dx As a result, the pulses that they produce are small and can be rejected by a discriminator setting In contrast, scintillators are solids and therefore their higher density (and often higher Z) leads to greater energy deposition and signal size 139 Chapter 20 Solutions � Problem 20.4 Proportional counting on the Α plateau The alpha plateau corresponds to where the pulse height is large enough to register a count for the alpha particle, but not large enough to record the energy deposition of electrons Since any background that could enter the gas through the detector wall will be either an electron or photon (which produces an electron in the gas), these pulses will not be seen because they are under the discrimator value Any background that is registered must be an alpha particle, and these can only reach the gas if they are emitted from the inside of the detector wall So the source of background will be natural radioactivity either in the gas as an impurity or in the detector wall � Problem 20.5 2.22 MeV background line Around concrete shields and other environmental sources that contain large amounts of hydrogen or H2 O, one often sees a 2.22 MeV background line in gamma spectrum The source of this line is neutron capture on hydrogen, or H �n, Γ� H, which has a Q value of 2.22 MeV The neutrons are primarily produced by cosmic ray interactions in the atmosphere and local environment � Problem 20.6 Coincidence vs Singles Counting Statistics We recall that the optimal division of time is given by Eqn 3.54 and yields a fractional uncertainty in the net source rate S of Ε= �1 �T� �S�B� � B For our data, the fractional uncertainty in the first measurements is: S Ε� �S � B� � �1 � T� B where we substitute T � 1, S � and B � S Ε � 2.99535 We look for the background B that gives the same fractional uncertainty in S: Ε� �S � B� � �1 � T� B where we substitute T � 1, S � B � S Ε � 2.99535 � � B 3�B Solving the above equation for B yields: B = 1.955 and therefore S=1.0045 The problem is also solvable using approximations Suppose we could assume S [...]... number of counts Ntotal � � r � ttotal Since for this problem, we are given and ttotal for each group, we can find Ntotal for each group Wtih N A and NB ( total counts from Group A and Group B), we want to find out if the difference between � r � A and � r �B is significant Define this difference as: �= NA TA - NB TB where T A is the total time for Group A (i.e., I * tI ), and similarly for Group... thickness in mfp units I N s Then: Σ y2 � � N � 1 0 where f � r= T0 T T s s 1 1 �� I T Ns 0 o �1 � e y T0 1 1 ��I T f Ts 0 �1 � e y f 1� f � Define the relative variance to be r, where: 1 1 I0 T fy2 �1 � e y f 1� f � Choosing I0 T � 1 for convenience: � 29 Chapter 3 Solutions r� y, f � � f �y 1� f �1 y2 f Let's plot the shape of the variance surface as a function of sample thickness y and fraction of time... for t < tc Looking at the v(t) expression for longer times, the voltage builds up to the maximum value of Q during C the collection time tc After tc , the differential equation is homogeneous and leads to an exponential fall off of the voltage This is the normal condition we want the maximum amplitude of the pulse is proportional to the total charge formed in the detector, and the total charge formed... % (or less) � Problem 4.8 Fano factor role in energy resolution 35 Chapter 4 Solutions Problem 4.8 Fano factor role in energy resolution The energy resolution is given by �E E � 2.355 ΣE E � � 2.355 F Σn n � 2.355 F n Given a resolution of 0.5%, F= 0.1, we solve for n F R � 2.355 n n� 5.5456 F R2 We then substitute F = 0.1 and R=0.005 in the last result to find the number of carriers n: n � 22 184 �... and the total amount of time allowed for the measurements to be done (20 min) to find the numerical values of Tsb and Tb sb Tsb � b Tb Tb � Tsb � Ttot The resulting solutions for Tsb and Tb solved from the equations above is thus: sb b Ttot Tsb � Tb � sb b �1 Ttot sb b �1 We substitute sb = 84.6, b = 7.3 and Ttot � 20 to get the numerical values of Tsb and Tb in minutes (i.e the optimal times for measuring... Chapter 3 Solutions 5 4 3 2 1 1 2 3 4 5 We can also use a nonlinear root finder to find the numeric value for the numerator, which yields a more accurate answer for the optimal wait time �t: �t � 2.22 Λp � Problem 3.22 Optimal choice of attenuation coefficient to minimize error in transmission thickness measurement a) First, we define the expression for the measured material thickness xm (different than... (counting efficiency), T is the count time in seconds, and f is the branching ratio for the observed radiation (the result is expressed in Bq) Α� ND ΕT f We substitute Ε = 0.15, T = 30!60 and f = 0.85 to get the minimum detectable amount (MDA) of Cs-137 on the filter Α � 1.12 Bq � Problem 3.25 Intervals between events (a) For a fixed frequency f, the period is T=1 /f, and the mean wait time is (1/2)T For an... probability of 0 counts must be less than 1% to satisfy this condition We again define our Poisson Distribution, but this time with a mean value of 2.87*t We define the probability of observing no counts to be 1%, and we solve for the resulting value for "t" that satisfies this condition: ��Μ Μk k� � 01 We substitute Μ=2.87t and k=0 then solve for t to get the minimum counting time required (in minutes) to ensure... we want to collect all of the charge, and have some time to measure this maximum voltage Although this is beyond the question being asked, let's solve the differential equation which describes the circuit of a capacitor and resistor in parallel, being fed a constant current of Q � tc during the collection time tc , from the detector This is done below, where "v(t)" is the voltage-time profile for the... settings is understood to be a value resulting from a defined procedure of counting the source and the background for the same length of time, which as you have seen, may not be optimal Nevertheless, it is a well accepted figureof-merit (a) LC is set to ensure that S=T-B gives only 5% false positives when S=0, We define "S" as the number of counts from the source alone, "T" is the number of counts including